The Genus Level of a Group

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1 The Genus Level of a Group Matthew Arbo, University of Oregon Krystin Benkowski, Marquette University Ben Coate, College of Idaho Hans Nordstrom, University of Portland Chris Peterson, Ponoma College Aaron Wootton, University of Portland November 5, 2008 Keywords: Automorphism Groups of Compact Riemann Surfaces Mathematical Subject Classification: 14H37, 30F20 Abstract We introduce the notion of the genus level of a group as a tool to help classify finite conformal group actions on compact Riemann surfaces. We classify all groups of genus level 1 and use our results to outline an algorithm to classify p-groups actions on compact Riemann surfaces. To illustrate our results, we provide a number of detailed examples. 1 Introduction The group of conformal automorphisms of a compact Riemann surface of genus σ 2 is always finite. An interesting problem is to determine all groups which can act on a surface X of genus σ 2 for a fixed σ. For low genus, due to great advances in computer algebra systems, this problem has been completed. Specifically, in [3], all groups which can act on surfaces up to genus 3 were classified. More recently, with the aid of the computer algebra system GAP, [6], the list of all groups which can act on surfaces of genus σ for genera 2 σ 48 were classified in [2] and stored in a database in GAP. Though in principle the algorithms developed to solve this problem could be used to determine automorphism groups of surfaces of higher genus, the computations become much more complicated, and complete classification does not seem reasonable. A different approach to classifying automorphism groups is to fix some other property than the genus of the surface X. One can, for instance, Partially supported by the NSF grant DMS The authors are grateful to Michael Sawdy of Rainier Middle School for his contributions to this work 1

2 restrict how a group may act on the surface, or one can restrict the types of groups under consideration which act on X. In many cases, complete classification results are possible. Such results are proliferant in the literature, see for example [1], [4], [7], [8] and [14]. One motivational reason for classifying automorphism groups by imposing further restrictions on the way an automorphism group can act on a surface is that it may aid in answering other questions about the surface. For example, in certain cases, the classical problem of determining a defining model for a compact Riemann surface X is possible when assumptions are made about the existence of automorphisms on the surface acting in a particular way, as done in [5] or [13]. Another motivational reason for classifying families of surfaces is that it may provide new techniques to the more general problem of classifying all groups which can act on surfaces of genus σ 2. Suppose G is a finite group of automorphisms of a compact Riemann surface X of genus σ 0. We define the genus level of G to be the number of distinct genera of the quotient spaces X/H where H runs over the non-trivial subgroups of G. In the following, we classify all groups which have genus level 1. This is an extension of [8] where full results are given for each group G which can act on a surface of genus σ 0 with the property that every quotient space X/H has genus 0 where H runs over all the non-trivial subgroups of G. We use our results to provide an inductive method to determine all p-groups which act as automorphism groups on surfaces of genus σ 2 (though it should be noted that with minor modifications, this algorithm would extend to a classification of non-simple groups). 2 Preliminaries Suppose that G is a group of conformal automorphisms of a compact Riemann surface X of genus σ 2 upon which we assume G acts faithfully. We need the following definitions: Definition 2.1. The signature of G is defined to be the tuple of integers (g; m 1, m 2,..., m r) where the orbit space X/G has genus g and the quotient map π G : X X/G is branched over r points with ramification indices m i for 1 i r. We say that the genus of G is g. Definition 2.2. We define the genus level of G to be the the number of distinct genera among those of the non-trivial subgroups of G. If G has genus level n, we define the genus level vector to be the n-tuple of increasing integers (t 1,..., t n) where t i is the genus of some non-trivial subgroup of G. Definition 2.3. A vector of elements (α 1,..., α t, β 1,..., β t, γ 1,..., γ r) is called a (t, m 1,..., m r)-generating vector of G if all of the following hold: 1. G = α 1,..., α t, β 1,..., β t, γ 1,..., γ r 2. O(γ i) = m i for 1 i r 3. Q t Qr [αi, βi] j=1 γj = e 2

3 We refer to the elements α i and β i as hyperbolic generators and the elements γ i as elliptical generators. The (t; m 1, m 2,..., m r)-generating vectors of a group G provide us with a way to determine when G acts on a surface of genus σ 2. Specifically, we have the following. Theorem 2.4. A group G acts with signature (t; m 1,..., m r) on a surface X with genus σ 2 if G has a (t; m 1,..., m r)-generating vector and the Riemann-Hurwitz formula holds: «2σ 2 = G 2t 2 + 1! 1mi Proof. For details, see for example [3]. If a group G acts on a surface X, then each of its subgroups also acts on X. The corresponding signatures of subgroups can be calculated from the signature of G using the following theorem of [10]. Theorem 2.5. Suppose G has signature (t; m 1,..., m r) and generating vector (α 1, β 1,..., α t, β t, γ 1,..., γ r). Suppose H G, set d = G / H and let θ : G S d be the map induced by permutation on the left cosets of H. Then the signature of H is where (s; n 11, n 12,..., n 1ρ1, n 21,... n 2ρ2,..., n r1..., n rρr ) (1) (i) θ (γ j) has precisely ρ j cycles of length m j m n j1,..., j n jρj, respectively; (ii) s satisfies 2s 2 + ρ X i j=1 1 1 «= d 2t 2 + n ij 1 1 m i «!. (2) Conversely, if d G and there exists a map θ : G S d and an integer s such that conditions (i) and (ii) hold, then there exists H G of index d and signature (1). In the special case that H is a normal subgroup of G, calculation of the signature is much easier. Theorem 2.6. Suppose G has signature (t; m 1, m 2,..., m k ) and H G. Then the signature of H is «s; m1,..., m1, m2,..., m2,..., mr,..., mr, n 1 n 2 n r n 1 a 1 times n 2 a 2 times n r a where O (ρ (γ i)) = n i under ρ : G G/H, a i = G/H /n i and s satisfies the following: 2s 2 = G H 2t n i «!. 3

4 Remark 2.7. Observe that we can use Theorems 2.5 and 2.6 to reconstruct the signature of G given the signature of H and the map θ : G G/H. The inconsistency of the notation for the order of the elliptic generators of H in Theorems 2.5 and 2.6 is to preserve the form of the summand in the Riemann-Hurwitz formula. To avoid confusion, we shall clearly distinguish between the normal and non-normal cases. Remark 2.8. It is an immediate consequence of Theorem 2.5 that t s. Note that this also implies, if G has genus vector (t 1,..., t r), then G must have genus t 1. All finite groups which can act on surfaces of genus 0 and 1 are well known, see for example [11] for genus 1 and [12] for genus 0. Most of the the results and definitions we have introduced for groups acting on surfaces of genus σ 2 hold for genus 0 and 1 too and will be relevant in our work. We summarize below. Theorem The possible finite groups and corresponding signatures for groups which can act on a surface of genus σ = 1 are the following: Group Signature C n C m (1; ) (C n C m ) C 2 (0; 2, 2, 2, 2) (C n C m ) C 3 (0; 3, 3, 3)) (C n C m ) C 4 (0; 2, 4, 4) (C n C m ) C 6 (0; 2, 3, 6) Table 1: Groups of Automorphisms of Genus 1 Surfaces We call these signatures toroidal signatures. 2. The possible finite groups and corresponding signatures for groups which can act on a surface of genus σ = 0 are the following: Group Signature C n (0; n, n) D n (0; 2, 2, n) A 4 (0; 2, 3, 3) S 4 (0; 2, 3, 4) A 5 (0; 2, 3, 5) Table 2: Groups of Automorphisms of Genus 0 Surfaces We call these signatures spherical signatures. We finish this section with an explicit example to illustrate the use of these results. 4

5 Example Consider the group Q 8 = x, y x 4, x 2 y 2, yxy 1 = x 1 The vector `x 2, x 2, x, x, y, y is a (0; 2, 2, 4, 4, 4, 4)-generating vector for Q 8. Using the Riemann-Hurwitz formula from Theroem 2.4, it follows that this is a group of automorphisms of a surface of genus σ = 1 + G ( 1) + G /2 6X (1 1/m i) 1 = «4 = 9. Consider the normal subgroup N = x and the canonical homomorphism ρ : G G/N. We have O `ρ `x 2 = 1, O (ρ (x)) = 1, and O (ρ (y)) = 2. By Theorem 2.5, it follows that H has signature (s; 2, 2, 2, 2, 2, 2, 4, 4, 4, 4) where 6X 1 s = 1+ G / N (g 1)+ G /2 N (1 1/n i) = «= Groups of Genus Level 1 We want to classify all genus level 1 group actions on all compact Riemann surfaces. In [8] all genus level 1 groups with genus level vector (0) are classified for groups acting on surfaces of genus σ 0. Using Theorem 2.9 we get the following classification for groups acting on genus 1 surfaces with genus level 1. Theorem 3.1. Suppose G, a finite group, acts on a surface of genus 1 and has genus vector (1). Then G = C n C m where C n and C m are finite cyclic groups. It remains to classify all genus level 1 groups acting on a surface of genus σ 2 with genus vector (t) for t 1. We start with the following important result. Proposition 3.2. Suppose G has signature (t; m 1, m 2,..., m r), where t 1. Let H be a nontrivial proper subgroup of G which also has genus t. Then t = 1 and the elliptical generators γ i H for all i. Proof. We set t = s in Equation 2 of Theorem 2.5 and rearrange, obtaining 2t 2 + ρ X i j=1 1 1 «n ij 2t 2 G G (2t 2) = H H «G 2 (1 t) H 1 = G G (2t 2) + H H = G H «1 1mi 1 1mi «1 1mi «ρ X i j=1 ρ X i j=1 1 1 «n ij 1 1 «n ij 5

6 Now, we have n ij m i, since n ij m i, and we know that ρ i G / H. Furthermore, equality will hold in each case exactly when θ (γ i) is the idenity permutation, consisting of G / H 1-cycles. Then «G 2 (1 t) H 1 = G H = 0 G H 1 1mi «1 1mi «ρ X i j=1 G H 1 1 «n ij «1 1mi Since t 1, the inequality cannot be satisfied unless t = 1. Thus each γ i acts as the identity permutation on the cosets of H, i.e., γ i H for all i. Remark 3.3. A consequence of this result is that if G has genus t > 1, then all proper subgroups have genus strictly greater than t. This allows us to classify all genus level 1 groups with genus level vector (t) for t > 1. Specifically, we have the following. Corollary 3.4. Suppose G acts on a surface of genus σ 2 and has genus level vector (t) for some t 2. Then G = C p where C p is cyclic of prime order p, the possible signatures of G are (t; p, p,..., p), where r 2 for all p and the additional condition that r is even when p = 2 and the genus σ satisfies σ = 1 + p (p 1)r (t 1). 2 2 An obvious but very useful consequence of this result is the following. Corollary 3.5. If G acts with signature (1; m 1, m 2,..., m r) and all γ i H for some subgroup H, then H has genus 1. The last case we need to consider is when a group, with genus level vector (1), acts on a surface of genus σ 2. Theorem 3.6. Suppose G acts on a surface of genus σ 2 and has genus level vector (1). Then G = C p n or G = Q 8. The possible signatures of C p n are (1; p, p,..., p), where r 2 for all p with the additional condition that r is even when p = 2, and the genus σ satisfies σ = 1 + pn 2 (t 1) pn 1 (p 1)r. 2 The possible signatures of Q 8 are (1; 2, 2,..., 2), where r is odd and the genus σ satisfies σ = 8t + 2r 7. Proof. Suppose that the generating vector of G is (α, β, γ 1,..., γ r). Since every non-trivial subgroup has genus 1, Corollary 3.5 implies that all γ i lie in each non-trivial subgroup of G. By assumption, m i 2, and r 1 6

7 since (1; ) does not satisfy the Riemann-Hurwitz formula for a group action on a surface of genus σ 2. It follows that the intersection of all non-trivial subgroups of G contains each γ i and hence is non-trivial, and thus G has a unique minimal non-trivial subgroup. The only groups which satisfy this condition are C p n and Q 2 n, [9, Theorem 5.3.6]. We now proceed by cases. D E First consider G = Q 2 n = x, y x 2n 1, x 2n 2 y 2, yxy 1 x. Let H = y2 G and let θ : G G/H denote the canonical quotient map. Observe that G/H = D 2 n 2, the dihedral group of order 2 n 1. By our remarks above, all γ i H. Now αβα 1 β 1 γ 1γ 2 γ r = e, so θ (e G) = θ `αβα 1 β 1 γ 1γ 2 γ r = [θ (α), θ (β)] = eg/h. It follows that θ (G) must be abelian, since it is generated by θ(α) and θ(β). The only abelian dihedral group is the Klein 4 group. It follows that G = Q 8. Now consider G = Q 8 = x, y x 4, x 2 y 2, yxy 1 x. Then G has a (1; 2, 2,..., 2)-generating vector `y, x, y 2, y 2,..., y 2 for any odd r. For r times even r, the generating vector must be of the form `α, β, y 2, y 2,..., y 2 for some α, β G. However, when r is even, the product of r copies of y 2 is the identity. It follows that α and β must commute and generate G, and no such generators of Q 8 exist, and hence r cannot be even. To finish, we need to show that every non-trivial subgroup of G has genus 1. First observe that the center Z(G) is the intersection of all non-trivial subgroups. By Remark 2.8, since we are assuming G has genus 1, it suffices to prove that Z(G) has genus 1. But this follows by Corollary 3.5. To determine σ, we simply applydtheorem 2.4. Finally, consider G = C p n = x x pne. Since C p n is abelian and all D γ i are in the minimal subgroup x pn 1E = a, the order of each γ i is p. Moreover, the commutator [α, β] will be equal to the identity, and hence γ 1... γ r = e. It follows that we must have r 2. Furthermore, if p = 2 then γ i = a for all i. In this case the product γ 1... γ r = e if and only if r is even. To finish, we need to construct generating vectors. For r even, the vector (x, x, a, a 1, a, a 1..., a, a 1 ) is a generating vector for G. For r odd, we know p 2 and consequently (x, x, a, a 1, a, a 1..., a, a 1, a, a, a 2 ) is a generating vector for G. To determine σ, we simply apply Theorem 2.4. Combing the results of [8], Theorems 3.1, 3.6 and Corollary 3.4, we get a complete classification of genus level 1 group actions. We summarize these results. Theorem 3.7. Suppose G is a finite group acting on a compact Riemann surface of genus σ 0 and G has genus level 1. Then we have the following possibilities: 1. If G has genus level vector (0) then G is either cyclic, generalized quaternion (Q 2 n), polyhedral or Zassenhaus metacyclic; 7

8 2. If G has genus level vector (1) and σ = 1, then G = C n C m, a direct product of two finite cyclic groups; 3. If G has genus level vector (1) and σ 2, then G = Q 8, the quaternion group, or G = C p n, a cyclic group of prime power order p n ; 4. If G has genus level vector (t) for t 2, then G = C p, a cyclic group of prime order p n. 4 Determining p-groups Suppose G is a non-simple group of automorphisms acting on a surface X of genus σ 0 with genus level vector (t 1,..., t n). If N is a normal subgroup G with genus t i, then G/N acts on the surface X/N of genus t i. The genus level vector of the action of G/N on X/N is clearly related to the genus level vector of the action of G of N. The precise relationship is is described by the following. Lemma 4.1. Suppose G acting on X has genus level vector (t 1,..., t k ) and N G is a non-trivial normal subgroup of G of genus t i. Then G/N has genus level vector (m 1,..., m l ) where the m j run over the genera of each subgroup K G with N K. Proof. Suppose G and N are as given. If K G/N, let K > N denote its pre-image in G given by the correspondence theorem. Then the quotient space (X/N)/( K) is naturally homeomorphic to the quotient space X/K. In particular, the genus of the quotient space (X/N)/( K) will be equal to the genus of the quotient space X/K. Conversely, using a similar argument, if N K, then the image of K in the quotient group G/N will have the same genus as K. The result follows. The following is a simple consequence of the relationship described in Lemma 4.1. Corollary 4.2. Suppose G has genus level vector (t 1,..., t k ) for k 2 and N G is a non-trivial normal subgroup of G of genus t i. Then unless G has genus level vector (0, 1) and t i = 1, the genus level vector of G/N is strictly shorter in length than the genus level vector of G. These observations suggest we can construct an inductive algorithm to help determine non-simple groups of automorphisms of compact Riemann surfaces dependent upon the length of the genus level vector. Specifically, if the genus level of a group G is at least two, then except under the exception provided above in Corollary 4.2, the genus level of a nontrivial quotient group will be strictly smaller than the genus level of a given group. This coupled with the classification of genus level 1 actions in Section 3 provides an inductive process for classification. Though in principle such an algorithm is possible, the calculations become difficult very quickly, especially the calculation of signatures. Therefore, rather than a general algorithim for all non-simple groups, we shall construct an algorithm for p-groups. Henceforth assume that G is a p-group for some prime p. Before we construct the algorithm, we shall develop several results specific to p-groups. 8

9 Theorem 4.3. Suppose that G is a p-group of order p n acting on a surface of genus σ with genus level vector V = (t 1,..., t k ) and C is a normal cyclic prime subgroup of G with genus t i, so C has signature (t i; p,..., p) where f times f = 2σ 2 2p(ti 1) p 1 Then the quotient group G/C acts on a surface of genus t i with genus level vector (t 1, t β1,..., t βs ) where β 1,..., β s run over the genera of each subgroup K with C K G. If the signature of G/C is (t 1; m 1,..., m r), then the signature of G is (t 1; a 1m 1, a 2m 2,... a rm r, p,..., p), l times where a 1,..., a r {1, p}, and f and l satsify f = lp n 1 + (a i 1)p n 1 (p 1)m i. Moreover, if V = (α 1,..., α t, β 1,..., β t, γ 1,..., γ r) is a generating vector for G, then a i = p if and only if C γ i and γ r+i C for all i 1. Proof. The specified action of G/C, the genus level vector, and generating vector are consequences of Lemma 4.1 and Theorems 2.4 and 2.6. To determine the signatures, we simply generalize Proposition 3 of [12]. As indicated by Corollary 4.2, the only barrier preventing us from constructing such an algorithm is the case where G has genus level vector (0, 1). The groups acting on a surface of genus σ = 1 with genus level vector (0, 1) are given in Theorem 2.9, so we only need to consider groups acting on surfaces of genus σ 2. Theorem 4.4. Suppose G is a group acting on a surface X of genus σ 2 with genus level vector (0, 1) with the property that if C is any normal cyclic prime subgroup, then C has genus 1 and G/C has genus vector (0, 1). Then one of the following completely describes G: 1. G = Q 2 n acts with signature (0; 2, 2,..., 2, 4, 4, 4, 4) and X has genus σ = 2 n 2 (r + 2) X has genus 3 and we have one of the following cases: (a) G = V 4 acts with signature (0; 2, 2, 2, 2, 2, 2) (b) G = C 4 C 2 acts with signature (0; 2, 2, 4, 4) (c) G = D 4 acts with signature (0; 2, 2, 4, 4) (d) G = D 4 acts with signature (0; 2, 2, 2, 2, 2) (e) G = C 4 C 4 acts with signature (0; 4, 4, 4) (f) G = D 2,8,5 acts with signature (0; 2, 8, 8) (using the same notation in [3]) 9

10 (g) G = (C 4 C 2) C 2 acts with signature (0; 2, 2, 2, 4) (h) G = D 2,8,5 C 2 acts with signature (0; 2, 4, 8) Proof. Fix a normal cyclic prime subgroup C G of genus 1. Since C has genus 1, the quotient space will have genus 1 and so by assumption, the group G/C will have genus vector (0, 1) and act on a surface of genus 1. It follows that we can only have p = 2 or p = 3. Before we determine G explicitly, we make some simple observations about the structure of G to determine all such groups which can act on surfaces of genus σ 3. First, we are assuming that both G and the quotient group G/C have genus vector (0, 1). It follows that G has genus 0, there is a subgroup K with genus 1 of order at least p 2 with C K, and consequently G must have order at least p 3. We shall examine K. In [3], all groups and signatures which can act on surfaces of genus 2 and 3 are classified. By simply checking these lists, we see that for genus 2 no such K exists, and for genus 3, no such K exists for p = 3. To determine all 2-groups acting on genus 3 surfaces with these properties, we simply proceed through the list in [3] to extract all the groups and signatures which satisfy the requirements. Henceforth, we shall assume that the genus of X is at least 4. We shall show that G has a unique cyclic prime subgroup. Suppose not and let H denote a cyclic prime subgroup different from C. Then the group L C, H is elementary abelian of order p 2. Note that by Theorem 3.6, the genus of L must be 0, so it follows that L will have signature (0; p, p,..., p) for some r. Let V denote a (0; p, p,..., p)-generating vector for L. Now if K L is any non-trivial proper subgroup, by assumption, the genus of K must be 0 or 1. Observe that the genus of K L is completely determined by the number of elements of V with non-trivial image under the quotient map ρ: L L/K. Specifically, if the genus of K is 0, precisely two generators will have non-trivial image. If the genus of K is 1 and p = 2, four will have non-trivial image; if p = 3, three will have non-trivial image. We consider the cases p = 2 and p = 3 separately. If p = 2, then 4 elements of V will have non-trivial image under ρ: L L/C. It follows that all remaining elements of V lie in C, and hence will have non-trivial image under ρ: L L/K for any other non-trivial subgroup K L. This can only happen if r = 6 and each non-trivial subgroup K L contains two elements of V, or r = 5 and two subgroups each contain 2 and the other contains 3. In each of these cases, we apply the Riemann-Hurwitz formula and find that either σ = 2 or σ = 3, counter to our assumption that σ 4. If p = 3, we use a similar argument. Specifically, the only possible signature for L is (0, 3, 3, 3, 3) where a (0, 3, 3, 3, 3)-generating vector for L has one element in each non-trivial subgroup K of L. However, it is easy to show that no such generating vector for L exists and hence no such group exists. We shall henceforth assume that G has a unique cyclic subgroup. Since G has a unique cyclic subgroup, it follows that either G = Q 2 n or G = C p n. We consider each of these cases. Suppose first that G = C p n and let K be the subgroup of maximal order of G that has genus 1. Then 10

11 the signature of K will be (1; p,..., p). Since G/K will have genus level vector (0) and act on a surface of genus 1, it must either be C 4, or cyclic of prime order 2 or 3. Using the toroidal signatures and Theorem 4.3, we can reconstruct the signature of G for each of these cases. If G/K is cyclic of order 3, then G will have signature (0; 3,..., 3, 3a, 3b, 3c) where a, b and c are either 1 or 3. Since G must be generated by elliptic generators, it follows that G can have order at most 9 contradicting the fact that G p 3. Likewise, if G/C is cyclic of order 2, then G will have signature (0; 2,..., 2, 2a, 2b, 2c, 2d) where a, b, c and d are either 1 or 2. It follows that G has order at most 4, again a contradiction. Finally, if G/K = C 4, then there would exist L with L > K and G/L = C 2 which we have already shown cannot happen. Now suppose that G = Q 2 n and fix the presentation D E x, y x 2n 1, x 2n 2 y 2, yxy 1 x for G. The group L = x is a cyclic subgroup of order 2 n 1 where n 3. By our previous observations, L must have genus 1. Similar to the last case, using the fact that G/L = C 2 and has genus level vector (0), we can reconstruct the signature for G. Specifically, the signature for G will be (0; 2, 2,..., 2, 2a, 2b, 2c, 2d) where a, b, c, and d are either 1 or 2. Any generating vector for G will be of the form (γ 1,..., γ r, γ r+1, γ r+2, γ r+3, γ r+4) where γ r+i / L for 1 i 4. However, if any γ r+i has order 2, then by uniqueness of C, it will lie in C and hence L. It follows that each γ r+i has order 4 and so the only possible signature for G is (0; 2, 2,..., 2, 4, 4, 4, 4). We finish by showing that this signature is admissible for every possible value of r. We know that the product γ 1γ 2 γ r is either e or y 2. We set γ r+1 = yx, γ r+2 = y 3 x, γ r+3 = y. If the product is e, we set γ r+4 = y 3 ; if it is y 2, we set γ r+4 = y. In either case, the generators multiply to e. Moreover, x = γ r+3γ r+2 and y = γ r+3 generate Q 2 n and thus, for each possible r, we have constructed a generating vector. Application of Theorem 2.6 shows, for this generating vector, all subgroups of L have genus 1 and the Riemann-Hurwitz formula gives the signature for X. Putting our work together, we can now outline the inductive method to determine all p-groups which can act on a compact Riemann surface dependent upon the genus level of the group. We note that all groups are known for genus 0 and 1 surfaces, so we restrict to surfaces of genus σ 2. Fix an n and assume that we know all p-group actions for all p-groups with genus level k n 1. To determine the possible p-groups actions, with genus level n, we do the following: 1. If n = 2, determine each possible p group G with genus level vector (0, 1) with the property that if C is any normal cyclic prime subgroup, then the group G/C has genus level vector (0, 1). This can be done using Theorem Else, any such group G of order p f admits a cyclic subgroup of order p such that G/C is a group of automorphisms of order p f 1 acting on a surface with a strictly shorter genus level vector than 11

12 the genus level vector of G. In particular, by assumption, we know the structure of every possible K = G/C. Therefore, to determine G, and its signature, we can do the following: (a) Determine the solutions of the short exact sequence 1 C G K 1 for each possible K. (b) For a given G and K from (a), each possible signature T of G can be determined from K using Theorem 4.3. Specifically, we run over all the possibilities by choosing a i = 1 or p for each i. (c) For a G and signature T from (b), we can determine all possible generating vectors and check, for each vector, that G has genus level n. If no valid generating vector exists, then there does not exist an action by G with this signature and with genus level n. 5 Examples We finish with some explicit examples to illustrate our results. Example 5.1. It is easy to determine all genus level 1 group actions of a cyclic prime group G of order p on a surface of genus σ 0. Specifically, the signature for such a group will be (t; p,..., p) where σ = 1 + p(t 1) + r(p 1). 2 For σ = 0 or 1, Theorem 2.9 gives all possible signatures. For σ 2, it is easy to show that a (t; p,..., p)-generating vector for G exists for all primes p, provided r 1 and when p = 2, r is even. The genus on which such a G acts is calculated using the Riemann-Hurwitz formula. Example 5.2. We can use our results and Example 5.1 to classify all group actions by cyclic groups of order p 2 for any prime p with genus level 2 on surfaces of genus σ 2. Let G be such a group. If C is the cyclic prime subgroup of G, Example 5.1 shows the signature of G/C is (t; p,..., p) with r 1 and if p = 2, then r is even. Using Theorem 4.3, since C γ for any non-trivial γ G, it follows that there are r elements of order p 2 in the signature of G. Therefore, the possible signatures of G are (t, p,..., p, p 2,..., p 2 ) r 1 times for certain values of r 1. Using Theorem 2.6, the genus level vector of such an action will be (t, t 1) where t 1 = 1 + p(t 1) + r(p 1). 2 12

13 We shall show that such a G exists for every possible r 1 and t. To show this, we need to show that there exists a (t, p,..., p, p 2,..., p 2 )-generating r 1 times vector for G. This can be done in a similar way to the final argument in Theorem 3.6. The genus on which such a G acts is calculated using the Riemann-Hurwitz formula. Example 5.3. We can use our results and the Example 5.1 to classify all elementary abelian group actions of order p 2 for any prime p with genus level 2 on surfaces of genus σ 2. Let G be such a group and let C be a cyclic prime subgroup of G. Using Example 5.1, the signature of G/C is (t; p,..., p). According to Theorem 4.3, the fact that all elements of G have order p implies the signature of G is (t, p,..., p, p,..., p) r 1 times for certain values of r 1. Using Theorem 2.6, the genus level vector of such an action is (t, t 1) where t 1 = 1 + p(t 1) + r(p 1). 2 To determine for which values of t and r 1 such a G exists, we need to construct generating vectors, or show that no such generators exists. We do this on a case by case basis. For the remainder of the proof, assume that G = x, y and let n = r+r 1 denote the number of elliptic generators. First, suppose that n = 0. Then t 2 since we are assuming σ 2. A (t; )-generating vector for G is the vector (x, y, 1,..., 1). Using Theorem 2.6, the genus of every possible non-trivial subgroup is t 1. It follows that such a G exists for every possible choice of t 2 with n = 0. Now suppose that the number of elliptic generators is n > 0 and t 0, 1. By assumption, G has genus level 2 and so by Proposition 3.2, every non-trivial subgroup of G must have genus t 1. Consider a generating vector (α 1,..., α t, β 1,..., β t, γ 1,..., γ n) for G. By Theorem 2.6, the genus of each non-trivial C G is completely determined by the number of γ i C. In particular, every non-trivial subgroup C G has the same genus provided each contains the same number of elliptical generators, γ i. Since there are p + 1 non-trivial subgroups of G, it follows that the only possible signature for G is (t, p,..., p) where n = r(p + 1)/p. For any such n-times r, and t 0, it is easy to construct a generating vector for G satisfying these properties by generalizing the final argument in Theorem 3.6. For example, if r is even and x 1,..., x p+1 is a set of non-trivial elements from distinct subgroups of G, then (1,..., 1, x 1, x 1 1,..., x 1, x 1 1,... x p+1, x 1 p+1,..., x p+1, x 1 p+1) 2t-times (r/2)-times (r/2)-times is a generating vector for G. It follows that such a G exists for every possible choice of t 2 provided n = r(p + 1)/p, and no such group exists otherwise. 13

14 The next case we consider is when t = 1. In this case, G could have subgroups with genus 1 as well as with genus t 1. If all the subgroups of G of order p have genus t 1, then it is easy to imitate the proof of the previous case to show that a generating vector for G exists only if n = r(p + 1)/p. Therefore, we only need consider the case when at least one of the subgroups of G of order p has genus 1. Let C G be a subgroup with genus 1. By Proposition 3.2, if (α 1,..., α t, β 1,..., β t, γ 1,..., γ n), then γ i C for all i. It follows that for any other subgroup K C, γ i / K for all i. Therefore, using Theorem 2.6, the genus of every subgroup K C of order p will be the same. Without loss of generality, suppose that C = x. It is easy to construct a generating vector for G satisfying these properties by generalizing the final argument in Theorem 3.6 and a generating vector always exists unless p = 2 and n is odd. The last case we need to consider is t = 0. As above, if every subgroup of order p has genus t 1, then a generating vector for G exists only if n = r(p + 1)/p. Therefore, we only need consider the case when at least one of the subgroups of G of order p has genus 0. In [14], all G which contain more than one cyclic prime subgroup with genus 0 were classified and there are only two possibilities for G. In the first case, G has signature (0; p, p, p) and generating vector (x, y, (xy) 1 ) and three genus 0 subgroups - the ones generated by x, y, and xy. Using Theorem 2.6, all other cyclic subgroups of G will have the same genus, and hence G has genus level 2. In the second case, G has signature (0; p, p, p, p) and generating vector (x, y, x 1, y 1 ) and two genus 0 subgroups - the ones generated by x and y. Using Theorem 2.6, all other cyclic subgroups of G will have the same genus, and hence G has genus level 2. Therefore, the last case we need to consider is when t = 0 and G contains a unique cyclic subgroup of genus 0. In this case, if C is the cyclic subgroup with genus 0 and (γ 1,..., γ n) is a generating vector for G, then by Theorem 4.3, γ i C for all but two values of i, say n 1 and n. As remarked previously, the only way two other subgroups C 1 and C 2 can have the same genus is if they contain the same number of γ i. This occurs exactly when G contains just two nontrivial, proper subgroups other than C, and each subgroup contains one of γ n 1 or γ n. Such restrictions force p = 2. Through our observations, the only possible generating vectors of G are of the form ( x,..., x, (xy), y). n 2-times Clearly this defines a generating vector when n is odd and does not for n even, and thus such a G exists only under these circumstances. References [1] R. Benim, Classification of Quasplatonic Abelian Groups and Signatures, Rose-Hulman Undergraduate Mathematics Journal, Vol. 9, Issue 1, 2008 [2] T. Breuer. Characters and Automorphism Groups of Compact Riemann Surfaces, Cambridge University Press (2001). [3] S. A. Broughton, Classifying Finite Group Actions on Surfaces of Low Genus, J. Pure and Appl. Alg., Vol. 69 (1990) pp

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